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(D) Let the initial velocity of the proton be $u$ and the final velocities of the proton and the particle be $v_1$ and $v_2$ respectively. Let the pro

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$m$

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(D) Let the initial velocity of the proton be $u$ and the final velocities of the proton and the particle be $v_1$ and $v_2$ respectively. Let the proton move at an angle $	heta$ with the initial direction, then the particle moves at an angle $(90^\circ - 	heta)$.By conservation of linear momentum:Along the initial direction ($x$-axis): $mu = mv_1 \cos 	heta + Mv_2 \cos(90^\circ - 	heta) = mv_1 \cos 	heta + Mv_2 \sin 	heta$ ...$(i)$Along the perpendicular direction ($y$-axis): $0 = mv_1 \sin 	heta - Mv_2 \sin(90^\circ - 	heta) = mv_1 \sin 	heta - Mv_2 \cos 	heta$ ...$(ii)$From $(ii)$, $Mv_2 \cos 	heta = mv_1 \sin 	heta$, so $Mv_2 = mv_1 	an 	heta$.Since the collision is elastic, kinetic energy is conserved:$\frac{1}{2}mu^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}Mv_2^2$$mu^2 = mv_1^2 + Mv_2^2$ ...$(iii)$Squaring and adding equations $(i)$ and $(ii)$:$(mu)^2 = (mv_1 \cos 	heta + Mv_2 \sin 	heta)^2 + (mv_1 \sin 	heta - Mv_2 \cos 	heta)^2$$m^2u^2 = m^2v_1^2 + M^2v_2^2$ ...$(iv)$Comparing $(iii)$ and $(iv)$:$m(mv_1^2 + Mv_2^2) = m^2v_1^2 + M^2v_2^2$$m^2v_1^2 + m M v_2^2 = m^2v_1^2 + M^2v_2^2$$m M v_2^2 = M^2v_2^2$$m = M$Thus, the mass of the unknown particle is $m$.

$A$ proton of mass $m$ collides elastically with a particle of unknown mass $M$ at rest. After the collision, the proton and the unknown particle are seen moving at an angle of $90^\circ$ with respect to each other. The mass of the unknown particle is:

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